Structure and Bonding in Organic Compounds, Chem 101-E0E

While organic is a vast subject, we will concern ourselves with one fundamental concepts that will form a solid foundation for those students who will be continuing with Chem 104 or Chem 150: an understanding of the structure and bonding in organic compounds.

This experiment is designed to acquaint you with structural aspects of organic chemistry through the assembly of molecular models of a variety of organic compounds. We will investigate structural aspects of a variety of hydrocarbons, as well as other functionalized (complicated) organic compounds (ones that contain oxygen in addition to carbon and hydrogen), and give you some additional practice in drawing formulas, predicting geometries, and identifying structural isomers for organic compounds.

Procedure

You and your lab partner should obtain one molecular model kit, which includes a variety of balls and sticks which are used to represent atoms and bonds, respectively.

Atoms Black Carbon Blue Nitrogen Red Oxygen Purple Halogens White Hydrogen Bonds Short, gray For single bonds (join two atoms with one short stick for a single bond; these will not bend) Long, gray For multiple bonds (join two atoms with two long sticks for a , 3 for a ; these will bend as needed)

Construct models of the organic compounds described below, and answer the questions associated with each model. You may find it useful to draw an expanded structural formula (or condensed structural formula) before you can start assembling a model for each .

1. Start with , the simplest alkane: CH4. Draw the expanded structural formula of methane below. Based on VSEPR theory, predict the geometry about the central C atom.

Structural formula Geometry

Now, build a model of methane, CH4. In the space below, draw, to the best of your ability, the three-dimensional structure of methane based on your model.

Structure of CH4

Note that the carbon atom in methane is tetrahedral (all of the H-C-H bond angles are ~109.5o), and as a result, the molecule is non-planar (all of the atoms do not reside in one plane). This is characteristic of the structures of all alkanes.

2. Now, consider the simplest , ethylene: C2H4. Draw the expanded structural formula of ethylene below. Based on VSEPR theory, predict the geometry about the two C atoms.

Structural formula Geometries

Now, build a model of ethylene, C2H4. In the space below, draw, to the best of your ability, the three-dimensional structure of ethylene based on your model.

Structure of C2H4

Does your model of ethylene confirm the geometries for each carbon atom that you predicted from VSEPR theory?

Note that as a result of the geometries of the carbon atoms in ethylene, the entire molecule is planar (all of the atoms do not reside in one plane), or flat. In addition, all of the bond angles for each carbon that participates in the double bond are ~120o. This is characteristic of the structures of all ; the atoms that participate in the double bond (i.e., the two carbons), and the atoms directly attached to those atoms (here, the four hydrogens), all occupy one plane, and the carbons are trigonal planar. 3. Now, consider another alkene, butene: C4H8. There are several structural isomers of this compound, and we will build models of all of them. Alkenes are hydrocarbons that contain at least one double bond between two carbon atoms; butene has one such double bond. In the spaces below, draw expanded structural formulas of two possible isomers of butene: one in which the double bond is between the first and the second carbon in a four-carbon chain (isomer “A”), and one in which the double bond is between the second and the third carbon is a four- carbon chain (isomer “B”). Complete the formulas by including hydrogens in the appropriate locations.

Isomer “A” Isomer “B”

Assemble models of these two isomers. After assembling them, try to move individual carbon atoms in each molecule by rotating around C-C (and C=C) bonds. In the two isomers, are all of the carbon atoms necessarily in the same plane all of the time?

Isomer “A”:

Isomer “B”:

These two isomers have different names. Isomer “A” is named 1-butene, because the double bond starts with carbon #1 in a 4-carbon chain. Isomer “B” is named 2-butene, since the double bond starts at carbon #2 in the 4-carbon chain.

Now, hold your model of 2-butene such that the double bond is straight up-and-down. Are the two –CH3 groups (called methyl groups) both to the left of the double bond, both to the right of the double bond, or one on each side?

Rebuild your model so that both methyl groups are on the same side of the double bond. This isomer of 2-butene is the cis isomer (the molecule is named cis-2-butene). Now, rearrange the atoms so that one methyl group is on each side of the double bond; this is the trans isomer (the molecule is trans-2-butene. Using your models as a guide, draw expanded structural formulas of the cis and trans isomers of 2-butene below.

Cis-2-butene Trans-2-butene There is another structural isomer of C4H10. Rearrange the atoms and bonds in one of your 2- butene models to identify this isomer. In the space below, draw an expanded structural formula of this fourth isomer of C4H10.

th 4 C4H10 isomer

The name of this fourth isomer of C4H10 is 2-methyl-1-propene.

4. Let’s consider some more complicated organic , specifically molecules that contain oxygen. The first group that we’ll consider will be alcohols, molecules that contain at least one -O-H group bonded to a single carbon. Using the models as a starting point, build two isomers of the alcohol C3H8O, named propanol. After building the models, draw expanded structural formulas of the two isomers you have built.

C3H8O isomer 1 C3H8O isomer 2

How do these two isomers differ?

We can use the same formula to illustrate another type of oxygen-containing organic compound, ethers, which are organic compounds in which the oxygen is bonded to two carbon atoms, not only to one as found in alcohols. Rearrange the atoms in one of your propanol models to construct an ether. Draw the expanded structural formula of this ether in the space below.

C3H8O ether 5. Finally, let’s examine two other groups of oxygen-containing organic molecules, aldehydes and ketones. Both of these groups of organic molecules contain a double bound between carbon and oxygen. Both of these types of molecules may be illustrated with the formula C3H6O. Build models of two isomers of C3H6O, both of which must contain the C=O unit. After you have built these models, draw expanded structural formulas for the compounds below.

C3H6O isomer 1 C3H6O isomer 2

In your two models, is the oxygen bonded to the first carbon in the 3-carbon chain, or the second carbon?

Isomer 1:

Isomer 2:

One of the compounds you have built is acetone (a component of fingernail polish remover), a ketone, and the other is propionaldehyde, an aldehyde. These two compounds are different because in one, the oxygen is doubly-bonded to the first (or last) carbon in the chain, while in the other, the oxygen is doubly-bonded to a carbon (somewhere) in the middle of the chain of carbon atoms. For your two isomers, which is the aldehyde, propionaldehyde, and which is the ketone, acetone?

Isomer 1:

Isomer 2: